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Modern Engineering Thermodynamics

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224 CHAPTER 7: Second Law of <strong>Thermodynamics</strong> and Entropy Transport and Production Mechanisms<br />

EXAMPLE 7.4 (Continued )<br />

Insulation<br />

FIGURE 7.12<br />

Example 7.4.<br />

m = 1.5 kg<br />

x 1 = 0<br />

T 1 = 20.°C<br />

p 1 = 0.10 MPa<br />

s 1 = ?<br />

State 1<br />

Constant volume process<br />

State 2<br />

s 2 − s 1 = ?<br />

m = 1.5 kg<br />

p 2 = 0.10 MPa<br />

∀ 2 =∀ 1<br />

s 2 = ?<br />

or, T 2 = T 1 . Therefore, the process must also be isothermal, and Eq. (7.33) gives the specific entropy change as<br />

s 2 − s 1 = c lnðT 2 =T 1 Þ = c lnð1Þ = 0<br />

Consequently, the entropy of an incompressible material is not altered by changing its pressure.<br />

EXAMPLE 7.5<br />

An apparatus contains 0.035 kg of air (an ideal gas). The apparatus is used to compress the air isentropically (i.e., at<br />

constant entropy) from a pressure of 0.100 MPa to a pressure of 5.00 MPa. If the initial temperature of the air is 20.0°C,<br />

determine the final temperature and specific volume of the air.<br />

Solution<br />

First, draw a sketch of the system (Figure 7.13).<br />

Air<br />

State 1<br />

m = 0.035 kg<br />

p 1 = 0.100 MPa<br />

T 1 = 20.0°C<br />

Isentropic process<br />

State 2<br />

Air<br />

m = 0.035 kg<br />

p 2 = 5.00 MPa<br />

T 2 = ?<br />

v 2 = ?<br />

FIGURE 7.13<br />

Example 7.5.<br />

The unknowns are the final temperature, T 2 , and specific volume, v 2 , of the air in the system. Since p 1 = 0.100 MPa,<br />

T 1 = 20.0°C, and p 2 = 5.00 MPa, then Eq. (7.38) can be used to find T 2 and v 2 as follows. From Thermodynamic Tables to<br />

accompany <strong>Modern</strong> <strong>Engineering</strong> <strong>Thermodynamics</strong>, Table C.13b, we find for air that k = 1.4; and solving Eq. (7.38) for T 2 gives<br />

k−1<br />

p<br />

<br />

T 2<br />

2 = T<br />

k<br />

5:00 MPa<br />

0:4<br />

1 = ð20:0 + 273:15 KÞ<br />

1:4 = 896 K = 623°C<br />

p 1<br />

0:100 MPa<br />

Using the ideal gas equation of state and Table C.13b for the gas constant of air (R air = 0.286 kJ/kg · K), we find the initial<br />

specific volume of the air to be<br />

v 1 = mRT 1 /p 1<br />

= ð0:035 kgÞð0:286 kJ=kg.KÞð20:0 + 273:15 KÞ/ð0:100 × 10 3 kN=m 2 Þ<br />

= 0:02934 m 3 =kg<br />

Then, solving Eq. (7.38) for v 2 gives<br />

1<br />

<br />

1 −<br />

T<br />

v 2 = v 2 1−k<br />

1 = ð0:02934 m 3 623 + 273:15<br />

=kgÞ<br />

0:4<br />

= 0:00180 m 3 =kg<br />

T 1<br />

20:0 + 273:15<br />

Exercises<br />

7. Determine the change in specific entropy as a 1.00 kg block of solid incompressible iron is heated from 20.0°C to 100.°C.<br />

(See Table 3.6 for specific heat values.) Answer: (s 2 − s 1 ) iron = 0.108 kJ/kg·K.<br />

8. Determine the change in specific entropy of air as it is heated from 20.0°C to 100.°C in a constant pressure (isobaric)<br />

process. Assume air behaves as an ideal gas. Answer: (s 2 − s 1 ) isobaric air = 0.242 kJ/kg·K.<br />

9. Determine the change in specific entropy of air as it is heated from 20.0°C to 100.°C in a constant volume (isochoric)<br />

process. Assume air behaves as an ideal gas. Answer: (s 2 − s 1 ) isochoric air = 0.173 kJ/kg·K.

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